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Frog Position After T Seconds

Difficulty: Hard

Problem Description

Given an undirected tree consisting of n vertices numbered from 1 to n. A frog starts jumping from vertex 1. In one second, the frog jumps from its current vertex to another unvisited vertex if they are directly connected. The frog cannot jump back to a visited vertex. If the frog can jump to several vertices, it jumps randomly to one of them with the same probability. Otherwise, when the frog cannot jump to any unvisited vertex, it jumps forever on the same vertex. Return the probability that after t seconds the frog is on the vertex target.

Key Insights

  • The problem involves traversing a tree structure using a graph traversal technique.
  • The frog can only jump to unvisited vertices, which requires keeping track of visited states.
  • A recursive or iterative approach can be employed to compute the probabilities at each time step.
  • The number of possible vertices to jump to is crucial for calculating the jump probabilities.

Space and Time Complexity

Time Complexity: O(n * t) - where n is the number of vertices and t is the time in seconds, as we might need to explore all vertices for each second. Space Complexity: O(n) - for storing the adjacency list of the tree and the visited states during traversal.

Solution

To solve the problem, we can use Depth-First Search (DFS) to explore the tree. We will maintain a probability value at each node, which represents the likelihood of the frog being at that node after t seconds. The algorithm follows these steps:

  1. Construct an adjacency list from the list of edges to represent the tree.
  2. Use a recursive DFS function that takes the current node, the remaining time, and the probability as parameters.
  3. At each node, check how many unvisited neighbors exist.
  4. If there are unvisited neighbors, recursively call the DFS function for each of those neighbors, updating the probability based on the number of ways to reach that neighbor.
  5. If the remaining time reaches zero, check if we are at the target node and return the probability accordingly.

Code Solutions

def frogPosition(n: int, edges: List[List[int]], t: int, target: int) -> float:
    from collections import defaultdict

    # Build the adjacency list for the tree
    graph = defaultdict(list)
    for u, v in edges:
        graph[u].append(v)
        graph[v].append(u)

    def dfs(node, time_left, prob, visited):
        if time_left == 0:
            return prob if node == target else 0
        
        visited.add(node)
        unvisited_neighbors = [neighbor for neighbor in graph[node] if neighbor not in visited]

        if not unvisited_neighbors:  # No unvisited neighbors
            return prob if node == target else 0

        # Probability to jump to each unvisited neighbor
        prob_per_neighbor = prob / len(unvisited_neighbors)
        total_prob = 0
        for neighbor in unvisited_neighbors:
            total_prob += dfs(neighbor, time_left - 1, prob_per_neighbor, visited)
        
        visited.remove(node)  # Backtrack
        return total_prob

    return dfs(1, t, 1.0, set())
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